题目链接：uva 1356 - Bridge

题目大意：在一座长度为B的桥上建若干个塔，塔的间距不能超过D，塔的高度为H，塔之间的绳索形成全等的抛物线。绳索的总长度为L。问在建最少塔的情况下，绳索的最下段离地面的高度。

解题思路：贪心的思想求出最少情况下建立的塔数。
二分高度，然后用积分求出两塔之间绳索的长度。

C++ 积分
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>

using namespace std;

double f (double a, double x) {
    double aa = a * a, xx = x * x;;
    return (x * sqrt(aa + xx) + aa * log(fabs(x + sqrt(aa + xx)))) / 2;
}

double parabola_length (double w, double h) {
    double a = 4 * h / (w * w);
    double b = 1.0 / (2 * a);
    return (f(b, w/2) - f(b, 0)) * 4 * a;
}

double bsearch (double l, double r, double d, double v) {
    while (r - l > 1e-5) {
        double mid = (r + l) / 2;
        if (parabola_length(d, mid) < v)
            l = mid;
        else
            r = mid;
    }
    return l;
}

int main () {
    int cas;
    scanf("%d", &cas);
    for (int kcas = 1; kcas <= cas; kcas++) {
        int D, H, B, L;
        scanf("%d%d%d%d", &D, &H, &B, &L);

        int n = (B + D - 1) / D;
        double d = B * 1.0 / n;
        double l = L * 1.0 / n;

        if (kcas > 1)
            printf("\n");
        printf("Case %d:\n%.2lf\n", kcas, (double)H - bsearch(0, H, d, l));
    }
    return 0;
}

C++ 辛普森
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>

using namespace std;
double A;

double f (double x) {
    return sqrt (1 + 4 * A * A * x * x);
}

double simpson (double a, double b) {
    double c = (a + b) / 2;
    return (f(a) + 4*f(c) + f(b)) * (b-a) / 6;
}

double asr (double a, double b, double eps, double S) {
    double c = (a + b) / 2;
    double L = simpson(a, c), R = simpson(c, b);
    if (fabs(L+R-S) <= eps * 15)
        return L + R + (L + R - S) / 15;
    return asr(a, c, eps/2, L) + asr(c, b, eps/2, R);
}

double asr (double a, double b, double eps) {
    return asr(a, b, eps, simpson(a, b));
}

double parabola_length (double w, double h) {
    A = 4 * h / (w * w);
    return asr(0, w / 2, 1e-5) * 2;
}

double bsearch (double l, double r, double d, double v) {
    while (r - l > 1e-5) {
        double mid = (r + l) / 2;
        if (parabola_length(d, mid) < v)
            l = mid;
        else
            r = mid;
    }
    return l;
}

int main () {
    int cas;
    scanf("%d", &cas);
    for (int kcas = 1; kcas <= cas; kcas++) {
        int D, H, B, L;
        scanf("%d%d%d%d", &D, &H, &B, &L);

        int n = (B + D - 1) / D;
        double d = B * 1.0 / n;
        double l = L * 1.0 / n;

        if (kcas > 1)
            printf("\n");
        printf("Case %d:\n%.2lf\n", kcas, (double)H - bsearch(0, H, d, l));
    }
    return 0;
}